Precalculus (6th Edition) Blitzer

The exact solution of the equation $2\sin 3x+\sqrt{3}=0$ on the interval $\left[ 0,2\pi \right)$ is $\frac{4\pi }{9}$ , $\frac{5\pi }{9}$ , $\frac{10\pi }{9}$ , $\frac{11\pi }{9}$ , $\frac{16\pi }{9}$ , and $\frac{17\pi }{9}$.
Let us consider the provided expression: $2\sin 3x+\sqrt{3}=0$ Add $-\sqrt{3}\text{ }$ on both sides: \begin{align} & 2\sin 3x+\sqrt{3}-\sqrt{3}=0-\sqrt{3} \\ & 2\sin 3x=-\sqrt{3} \end{align} And divide by $2$ on both sides: \begin{align} & \frac{2\sin 3x}{2}=\frac{-\sqrt{3}}{2} \\ & \sin 3x=\frac{-\sqrt{3}}{2} \end{align} On, the interval $\left[ 0,2\pi \right),$ the sine function is $\frac{-\sqrt{3}}{2}$ at $\frac{4\pi }{3}\text{ and }\frac{5\pi }{3}$. So, \begin{align} & \sin 3x=\frac{-\sqrt{3}}{2} \\ & 3x=\frac{4\pi }{3} \end{align} Or, \begin{align} & \sin 3x=\frac{-\sqrt{3}}{2} \\ & 3x=\frac{5\pi }{3} \end{align} Since the period of the sine function is $2\pi$, the general solution of the equation is: \begin{align} & 3x=\frac{4\pi }{3}+2n\pi \\ & x=\frac{4\pi }{9}+\frac{2n\pi }{3} \end{align} Or, \begin{align} & 3x=\frac{5\pi }{3}+2n\pi \\ & x=\frac{5\pi }{9}+\frac{2n\pi }{3} \end{align} To get different solutions, put $\text{ }n=0,1,2,3\ldots$ When $n=0$ , $x=\frac{4\pi }{9}\text{ or }x=\frac{5\pi }{9}$ When $n=1$ , \begin{align} & x=\frac{4\pi }{9}+\frac{2\pi }{3} \\ & =\frac{10\pi }{9} \end{align} Or, \begin{align} & x=\frac{5\pi }{9}+\frac{2\pi }{3} \\ & =\frac{11\pi }{9} \end{align} When $n=2$ , \begin{align} & x=\frac{4\pi }{9}+\frac{4\pi }{3} \\ & =\frac{16\pi }{9} \end{align} Or, \begin{align} & x=\frac{5\pi }{9}+\frac{4\pi }{3} \\ & =\frac{17\pi }{9} \end{align} Put the other values of $n$; then the solution becomes outside the interval $\left[ 0,2\pi \right).$ Thus, the exact solution of the equation $2\sin 3x+\sqrt{3}=0$ on the interval $\left[ 0,2\pi \right)$ is $\frac{4\pi }{9}$ , $\frac{5\pi }{9}$ , $\frac{10\pi }{9}$ , $\frac{11\pi }{9}$ , $\frac{16\pi }{9}$ , and $\frac{17\pi }{9}$.